利用openpyxl拆分多个工作表的工作簿

实现按目录拆分工作簿，源数据如下图

按目录拆分成N个文件。

上代码，没有找是否有整个sheet 复制的，先逐个cell复制解决问题。：

# encoding: utf-8
"""
@author: 陈年椰子
@contact: [email protected]
@version: 1.0
@file: Split_Xls.py
@time: 2019/9/24 0028 15:04

说明
"""
def Split_Xls(xls_file):
    from openpyxl import load_workbook
    from openpyxl import Workbook

    wb = load_workbook(xls_file)
    sheet_list = wb.sheetnames
    print(sheet_list)
    a_sheet = wb['目录']
    for i in range(3,6):
        sheet_name = a_sheet['B{}'.format(i)].value
        if sheet_name is None:
            break
        if sheet_name == '':
            break
        sr_sheet = wb[sheet_name]
        new_file_name = "{}.xlsx".format(sheet_name)
        print(sheet_name)
        wb_tg = Workbook()
        ws = wb_tg.active
        ws.title = sheet_name

        # 两个for循环遍历整个excel的单元格内容
        for i, row in enumerate(sr_sheet.iter_rows()):
            for j, cell in enumerate(row):
                # print(i,j,cell.value)
                ws.cell(row=i + 1, column=j + 1, value=cell.value)
        wb_tg.save(new_file_name)
        wb_tg.close()
    wb.close()



def Split_Xls2(xls_file):
    # 这个是通过删除其他的工作表，只留下要保存的工作表，这样就可以整个表复制，包括样式，过程曲折，但能达到效果。
    from openpyxl import load_workbook
    wb = load_workbook(xls_file)
    sheet_list = wb.sheetnames
    print(sheet_list)
    work_list = []
    a_sheet = wb['目录']
    for i in range(3,6):
        sheet_name = a_sheet['B{}'.format(i)].value
        if sheet_name is None:
            break
        if sheet_name == '':
            break
        work_list.append(sheet_name)
    wb.close()

    for sheet_name in work_list:
        new_file_name = "{}.xlsx".format(sheet_name)
        print('处理工作表', sheet_name, '\t保存文件', new_file_name)
        wb = load_workbook(xls_file)
        # print(wb.sheetnames)
        for del_sheet in sheet_list:
            if del_sheet != sheet_name:
                # print('del',del_sheet)
                wb.remove(wb[del_sheet])
        wb.save(new_file_name)
        wb.close()


Split_Xls2('test.xlsx')